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The correct set of four quantum numbers for the valence electrons of Cesium atom
(Z = 55) is :

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6, 1, 1, $+ \frac{1}{2}$

6, 0, 1, $+ \frac{1}{2}$

6, 0, 0 , $+ \frac{1}{2}$

6,1, 0 , $+ \frac{1}{2}$

answer is A.

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Detailed Solution

$_{37} R b : 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{10}, 4 s^{2} 4 p^{6}, 4 s^{1}$ Valence electron is in 5s orbital i.e.., $5 s^{1}$
$∴ n = 5, l = 0, m = 0, S = + \frac{1}{2} o r - \frac{1}{2}$

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